Question

measurements show that the enthalpy of a mixture of gaseous reactants decreases by 150. kj during a certain chemical reaction, which is carried out at a constant pressure. furthermore, by carefully monitoring the volume change it is determined that 102. kj of work is done on the mixture during the reaction.
calculate the change in energy of the gas mixture during the reaction. round your answer to 2 significant digits.
is the reaction exothermic or endothermic?
options: exothermic, endothermic

Explanation:

Step1: Recall the first law of thermodynamics

The formula is $\Delta E = q + w$, where $q = \Delta H$ at constant pressure.

Step2: Assign values to variables

$\Delta H = -150\ \text{kJ}$ (enthalpy decreases), $w = +102\ \text{kJ}$ (work done on the system).

Step3: Calculate $\Delta E$

$\Delta E = -150 + 102 = -48\ \text{kJ}$.

Step4: Determine reaction type

Negative $\Delta H$ means heat is released, so exothermic.

Answer:

• Change in energy: -48 kJ
• Reaction type: exothermic